{ "schema_version": "ocme-reviewed-question-reader-v1", "question_id": "q-20261015-advance-20261118-r07-pumping-lemma-regular-languages-0049", "pilot_number": "R0053-27374", "number": 27374, "family_id": "qf-mod49-pump-dfa-loop-witness", "title": { "zh-Hant": "第 27374 題：正規語言Pumping引理", "en": "Question 27374: A Complete Unary DFA Loop Gives an Invariant for Every Pump Count" }, "original_question": { "schema_version": "ocme-question-v0.1", "id": "q-20261015-advance-20261118-r07-pumping-lemma-regular-languages-0049", "batch_id": "batch-20261015-advance-20261118-r07-pumping-lemma-regular-languages-foundations", "language": "zh-Hant", "source_mko_ids": [ "mko-pumping-lemma-regular-languages" ], "title_zh": "正規語言Pumping引理", "stem_zh": "exists p forall w exists split forall i量詞與split constraints責任。操作=pump-dfa-loop-witness；資料={\"modulus\":24,\"word_length\":96}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。", "answer": { "kind": "expression", "value": "(24, 96, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 0], [0, 24, 0], 0, 24, 72, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, 是)", "explanation_zh": "Wave6第十二層核對=(24, 96, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 0], [0, 24, 0], 0, 24, 72, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, 是)；依完整明示資料獨立重算；positive witness由完整DFA loop支撐全部i，a^p b^p反例涵蓋全部合法splits；pumping property只必要不充分；單一split或bounded words不能證regular。" }, "difficulty": { "band": "undergraduate_core", "level": 3, "skills": [ "正規語言Pumping引理" ] }, "variant": { "family_id": "qf-mod49-pump-dfa-loop-witness", "seed": 202611187049, "parameters": { "modulus": 24, "word_length": 96, "domain": "Exact_regular_pumping_lemma", "operation": "pump-dfa-loop-witness", "semantic_template": "pump-dfa-loop-witness|{modulus:24,word_length:96}" } }, "verification": { "status": "candidate", "checks": [ "producer_exact_finite_or_symbolic_model", "producer_source_snapshot_check", "producer_substantive_template_deduplication", "producer_json_schema" ], "warning_zh": "候選須主端重算；不構成一般limit theorem、coverage、number-theory或language theorem證明、人工審定、Lean或Evidence。" }, "provenance": { "generator_role": "question_producer", "generator_model": "model_unknown", "generated_at": "2026-10-15", "contract_version": "ocme-question-factory-v0.1" } }, "original_english_edition": { "schema_version": "ocme-question-localization-v0.1", "question_id": "q-20261015-advance-20261118-r07-pumping-lemma-regular-languages-0049", "batch_id": "batch-20261015-advance-20261118-r07-pumping-lemma-regular-languages-foundations", "locale": "en", "translation_state": "ai_candidate", "source": { "question_sha256": "b2e72265fe815ba60d1d13b6836fb8149dc96db1ac52def3556f20369cb9cf12", "family_id": "qf-mod49-pump-dfa-loop-witness", "seed": 202611187049, "semantic_template": "pump-dfa-loop-witness|{modulus:24,word_length:96}" }, "fields": { "title": "A Complete Unary DFA Loop Gives an Invariant for Every Pump Count", "stem": "Use the unary DFA for lengths divisible by modulus=24, with accepted word length=96. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).", "answer": { "kind": "expression", "value": "(24, 96, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 0], [0, 24, 0], 0, 24, 72, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)", "explanation": "Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (24, 96, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 0], [0, 24, 0], 0, 24, 72, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)." }, "skills": [ "A Complete Unary DFA Loop Gives an Invariant for Every Pump Count" ], "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review." } }, "givens": [ { "id": "G001", "text": { "zh-Hant": "exists p forall w exists split forall i量詞與split constraints責任。操作=pump-dfa-loop-witness；資料={\"modulus\":24,\"word_length\":96}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。", "en": "Use the unary DFA for lengths divisible by modulus=24, with accepted word length=96. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified)." } } ], "steps": [ { "id": "S001", "uses": [ "G001" ], "before": "source roles = [pump-dfa-loop-witness]", "after": "derived = [[modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified]]", "relation": "implies", "rule_id": "source_ordered_fields_without_JS", "conditions": [], "explanation": { "zh-Hant": "此頁直接列完整來源欄序與mode特定角色，不靠JavaScript或由答案外形猜介面；No、undefined、infinity各保原意。", "en": "List the complete source field order and mode-specific roles without JavaScript or answer-shape inference;No,undefined and infinity retain their distinct meanings." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。", "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct." } }, "arithmetic": null }, { "id": "S002", "uses": [ "G001", "S001" ], "before": "Pumping ORIGINAL unary z length := sub(96, 24)", "after": "Pumping ORIGINAL unary z length = 72", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。", "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Pumping ORIGINAL unary z length，不以小數近似替代。", "en": "Recompute Pumping ORIGINAL unary z length with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "sub", "inputs": [ "96", "24" ], "output": "72", "label": "Pumping ORIGINAL unary z length" } }, { "id": "S003", "uses": [ "G001", "S002" ], "before": "source roles = [24, 96]", "after": "derived = [[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 17], [17, 18], [18, 19], [19, 20], [20, 21], [21, 22], [22, 23], [23, 0]], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 0], [0, 24, 0], 0, 24, 72]", "relation": "implies", "rule_id": "Pumping_complete_unary_DFA_cycle_and_actual_accepted_length", "conditions": [], "explanation": { "zh-Hant": "DFA狀態是長度mod m，逐每個狀態的unary transition加1取mod；前m+1狀態在位置0,m同為0。原word length真能被m整除，故x空、y長m、z為餘段給合法且接受的w。", "en": "The DFA state is length mod m;every unary transition adds1 modulo m.The first m+1 states repeat0 at positions0,m.The ORIGINAL word length is divisible by m,so x empty,y length m and the remaining z form a legal accepted word." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。", "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct." } }, "arithmetic": null }, { "id": "S004", "uses": [ "G001", "S003" ], "before": "source roles = [24, 96, 72]", "after": "derived = [EVERY integer i>=0: pumped length=i*modulus+z, z is divisible by modulus, state remains0 after each loop;accepted, 是]", "relation": "implies", "rule_id": "Pumping_ALL_i_loop_invariant_NOT_finite_pump_tests", "conditions": [], "explanation": { "zh-Hant": "任意i>=0重走同一完整loop i次回0，再讀z；因z亦為m倍數，終態仍0。以完整transition與全稱不變量支持所有i，不是只測若干pump次數。", "en": "For arbitrary i>=0,traverse the same complete loop i times and then z.Since z is also divisible by m,the final state remains0.Full transitions and this quantified invariant support EVERY i,not sampled pump counts." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。", "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct." } }, "arithmetic": null } ], "math_model": { "result": [ 24, 96, [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 0 ], [ 0, 24, 0 ], 0, 24, 72, "for_all_i_length=(i*modulus)+z_is_0_mod_modulus", true ] }, "conclusion": { "canonical_value": "(24, 96, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 0], [0, 24, 0], 0, 24, 72, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, 是)", "explanation": { "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。", "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged." } }, "definition_note": { "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。", "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate." }, "source": { "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db", "question_binding": { "batch_id": "batch-20261015-advance-20261118-r07-pumping-lemma-regular-languages-foundations", "batch_path": "public/data/questions/batches/batch-20261015-advance-20261118-r07-pumping-lemma-regular-languages-foundations/questions.jsonl", "batch_sha256": "657cec46a009582d6773652bd01d9e8b1a552cf28a6e0706bdf7b5813871f2ec", "original_line": 49, "original_line_sha256": "fa4f64df29c63b6cb784d59c0e889c5bfc43c9259d1005fafe8c084789456e06", "question_compact_sha256": "b2e72265fe815ba60d1d13b6836fb8149dc96db1ac52def3556f20369cb9cf12", "english_row_compact_sha256": "2bb7edd084898296b8c2e037f71dd7cfd22fa92e2b13af5931f04d7779e9a651", "english_manifest_path": "public/data/i18n/editions/batch-20261015-advance-20261118-r07-pumping-lemma-regular-languages-foundations.en.json", "english_rows_path": "public/data/i18n/questions/batch-20261015-advance-20261118-r07-pumping-lemma-regular-languages-foundations.en.jsonl" }, "mko_bindings": { "mko-pumping-lemma-regular-languages": { "sha256": "7466612ae549a8cda880e2921d8c9163ab7681371f543eb905240583b66e6ed4", "english_sha256": "b98a8e7b0fb61714b4bea18947593dabe418b1c724d8240eaba1b40c826bd9ec", "original_path": "public/data/mko/pumping-lemma-regular-languages.json", "english_path": "public/data/i18n/editions/mko-pumping-lemma-regular-languages.en.json" } }, "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff", "selected_part_sha256": "a54409c0bc3f68730e1072c6d3d772970462c1bc7ffc203b4030a6136a610283", "original_candidate_sha256": "aa30df68865ed41b03bd689f6efe154d451dc08722734f536d61aebc6de314c4", "original_candidate_line_sha256": "aa30df68865ed41b03bd689f6efe154d451dc08722734f536d61aebc6de314c4" }, "historical_source_state": { "review_state": { "derivation": "candidate_authored_and_locally_checked", "source_question_state": "candidate", "independent_AI_or_human_review": "NOT_PERFORMED", "formalization": "not_formalized", "Lean": "NOT_RUN" }, "scope_note": { "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。", "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged." }, "label": "Frozen producer metadata; current per-question review is separate below." }, "revision": { "id": "review-20261026-010-p54", "review_date": "2026-10-26", "owner_authorized_reader_publication": true, "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.", "review_receipts": { "semantic_review_sha256": "1b7a62514347b82518305d5feb4369a52d7d44c442a0f2b131e9eb3cd7b45f62", "question_first_witness_sha256": "79dc2a1077081f3ea2a03f75e32778163bf05e814cdf4c29ed86e567295f9ee1" }, "status": "second_context_reviewed", "mathematical_semantics": true, "bilingual_semantics": true, "steps_reviewed": 4, "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction", "exact_question_first_check": { "model_sha256": "43535b2d18359252aead8e3c7a215f45ce1755aed89737e831bd6cd76e000c29", "arithmetic_steps": 1 }, "uses_answer_as_premise": false, "human_review": "NOT_PERFORMED", "formalization": "not_formalized", "Lean": "NOT_RUN", "Canonical_truth_promotion": false, "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged." } }